atof

I have been looking into a problem whereby I am converting a float to a human readable format, and back. Namely a string. I have ran into issues using stringstream and found that atof produces "better" results. Notice, I do not print out the data in this case, I used the debugger to retrieve the values: const char *val = "73.31"; std::stringstream ss; ss << val << '\0'; float floatVal = 0.0f; ss >> floatVal; //VALUE IS 73.3100052 floatVal = atof(val); //VALUE IS 73.3099976 There is probably a reasonable explanation to this. If anybody can enlighten me I'd be greatful :). Answer is based on the

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Say, I have a (German) expression which reads 10.401,40 (in Mio EUR) , I'd like to convert this to a real float (in this case around 10 billions) in Python. This is what I have thus far: import re, locale from locale import * locale.setlocale(locale.LC_ALL, 'de_DE') string = "10.401,40 (in Mio EUR)" m = re.search(r'([\d.,]+)', string) if m is not None: number = atof(m.group(1)) * 10**6 However, it raises a ValueError ( ValueError: invalid literal for float(): 10.401.40 ). Why? Isn't the .setlocale() directive supposed to be handling exactly

本篇博客用于记录我自己用C++实现的一个计算器，目标是完成加减乘除带括号的四则运算，并在后期用工厂设计模式加以优化。 Part 1：calculate 1+1=2 实现这样的一个式子的计算，只需要用到字符串分割即可，一开始尝试了stringstream去先读入一整个字符串"1+2",然后创建了两个临时变量int和一个char，用>>去读入，但是发现读入的char放在中间被忽略掉了 string s ="12+34"; // stringstream ss(s); // int n1,n2,c; // ss>>n1>>c>>n2;//c is 34 ,the '+' is ignored 然后我就换用了substr去进行分割，substr有两个参数，一个是开始分出字串的位置，另一个是字串的长度，第二个参数默认值是npos，也就是字符串末尾。 int i =0; while(isdigit(s[i])) { ++i;//pos } // int n =atof("22.0"); // cout<<n; //float d = atof(s.substr(0,i).c_str()); cout<<cal( atof(s.substr(0,i).c_str()) , atof(s.substr(i+1).c_str()) ,s[i])<<endl;　　　　//string to const

c - Not including stdlib.h does not produce any compiler error! For historical reasons -- specifically, compatibility with very old C programs (pre-C89) -- using a function without having declared it first only provokes a warning from GCC, not an error. But the return type of such a function is assumed to be int, not double, which is why the program executes incorrectly. If you use -Wall on the command line, you get a diagnostic: $ gcc -Wall test.c test.c: In function ‘main’: test.c:5: warning: implicit declaration of function ‘atoi’ test.c:6: warning: implicit declaration of function ‘atof’

I am trying to convert a character array into double in c using atof and receiving ambiguous output. printf("%lf\n",atof("5")); prints 262144.000000 I am stunned. Can somebody explain me where am I going wrong? Make sure you have included the headers for both atof and printf. Without prototypes the compiler will assume they return int values. When that happens the results are undefined, since that doesn't match atof's actual return type of double . #include <stdio.h> #include <stdlib.h> No prototypes $ cat test.c int main(void) { printf("%lf\n", atof("5")); return 0; } $ gcc -Wall -o test test